# What is a nonscalar curvature singularity?

1. Aug 25, 2015

### pervect

Staff Emeritus
What is a nonscalar curvature singularity, in the context of "the https://en.wikipedia.org/w/index.php?title=Wave_of_death&action=edit&redlink=1 [Broken] is a gravitational plane wave exhibiting a strong nonscalar null https://en.wikipedia.org/w/index.php?title=Curvature_singularity&action=edit&redlink=1 [Broken], which propagates through an initially flat spacetime".

The quote is from wiki, https://en.wikipedia.org/w/index.php?title=Pp-wave_spacetime&oldid=666287121#Examples. There is a rather old post on the topic of "the wave of death", https://www.physicsforums.com/threads/wave-of-death.93654/, on PF which was quite interesting, but I'm not quite following the details as I'm not sure what a nonscalar curvature singularity is.

My best guess is that because scalar invariants, such as the Ricci scalar, all vanish for the pp wave class of space-times, one looks to higher-order tensors with "strong" components. What qualifies a component as being "strong" or "singular" is unclear to me at this point, I would speculate perhaps a tensor with dirac-delta function comonents would qualify?

Last edited by a moderator: May 7, 2017
2. Aug 26, 2015

### Staff: Mentor

My guess is exactly the same as yours. In a local inertial frame I think that you have a curvature tensor with at least two Dirac delta components such that they cancel each other and you wind up with a zero curvature scalar.

I don't know if it is the Riemann or Ricci curvature tensor, but my guess is that either would qualify.

3. Aug 27, 2015

### Ben Niehoff

The words "nonscalar curvature singularity" appear in Griffiths & Podolsky, but they are not defined. Have you looked in Hawking and Ellis? I don't own a copy.

4. Aug 27, 2015

### martinbn

Why shouldn't the singularity mean the usual geodesic (or some variant) incompleteness? And being nonscalar as you said, all scalars are well behaved.

5. Aug 27, 2015

### pervect

Staff Emeritus
I don't have Hawking and Ellis, alas, but it sounds like a good suggestion as to where to look - thanks.

Geodesic incompleteness (of timelike or null curves) makes sense in this context and I suspect it's what is meant. The points on the manifold where the metric isn't continuously differentiable would make the space-time geodesically incomplete if they are removed. Wald, after noting that singularities are hard to define, does mention geodesic incompleteness as the most satisfactory definition, and also the definition used in the various singularity theorems, which is good enough confirmation for me at this point.

Taking this all into account, then, a (relatively) simple example of a "wave of death" should be the aichelburg-sexl ultraboost, https://en.wikipedia.org/wiki/Aichelburg–Sexl_ultraboost.

My understanding of the ultraboost as a limiting sequence of a relativistic flyby, though, suggests that any particular test object (say a baseball) wouldn't necessarily be destroyed by the ultraboost - the "wave of death" isn't necessarily deadly. In the case of the ultraboost, for instance, it would be more like the baseball being given an impulse, as if it were hit by a bat. A glass ornament might well shatter under this treatment, the baseball might survive - depending on how hard the hit was. I think this also matches up reasonably well with the old PF post on the "wave of death".

It does seem a bit odd to talk about being able to predict the path (and fate) of the baseball when we just got through saying that the space-time was geodesically incomplete, but from a physical point of view (if not a mathematical one), it seems to make sense.

Last edited: Aug 27, 2015
6. Aug 27, 2015

### martinbn

Very interesting, do you know if any of the references at the end of the wiki page have more details about this spacetime?

Why would the incompleteness be a problem. The usual cosmological models are geodesically incomplete, but it is not a problem to find the world line of a particle (I assume that's what the path and fate mean), it's just not going to be very long.

7. Aug 27, 2015

### Ben Niehoff

Ah, ok, this actually does sound familiar. I had a homework problem a few years ago in my GR course based on a plane-wave version of this. A very simple metric is

$$ds^2 = - 2 du \, dv + a(u)^2 \, dx^2 + b(u)^2 \, dy^2.$$
I invite you to work out the Riemann and Ricci tensors for such a simple metric, it took me about half a page. The only nonvanishing component of the Ricci tensor is

$$R_{uu} = - \frac{a''}{a} - \frac{b''}{b},$$
which will impose one constraint on $a,b$ from the Einstein equation (so there is one free function left: an arbitrary wave profile). Both the Ricci scalar and the Kretschmann invariant vanish. Hence "no scalars".

The real fun happens if you set

$$a(u) = 1 + \lambda u \Theta(u), \qquad b(u) = 1 - \lambda u \Theta(u),$$
where $\lambda$ is some constant and $\Theta(u)$ is the Heaviside step function. Then the above metric represents an impulsive plane wave, which smacks every object in its path with a momentary, infinite force as it passes. It also focuses worldlines into caustics, causing objects to collide after they have been smacked.

8. Aug 28, 2015

### martinbn

How did you do the calculation on half a page? Are you doing something smart? It took me more than half a page just to find the connection 1-forms and the curvature 2-forms. Anyway I also must have made a mistake because the xx and the yy components of Ricci turned out to be non-zero. The only consolation is that the uu component came right. If I have the time I'll check my work and go to the second (fun) part of this exercise.

9. Aug 28, 2015

### Ben Niehoff

I did it in a coordinate basis, and the "trick" is predicting in advance which terms have to be zero, and using symmetry under the exchange of x and y along with a(u) and b(u).

"Half a page" might be an exaggeration. I forgot that the Christoffel symbols were done already. Altogether, I think it's 3/4 of a page.